Question: The equation of a circle $C$ is $x^2+y^2+6x-2y+1 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2+6x) + (y^2-2y) = -1$ $(x^2+6x+9) + (y^2-2y+1) = -1 + 9 + 1$ $(x+3)^{2} + (y-1)^{2} = 9 = 3^2$ Thus, $(h, k) = (-3, 1)$ and $r = 3$.